How do you factor the expression #4q^2 + 27r^4#?

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Answer 1 · 2016-03-25T00:06:55

#4q^2+27r^4=(2q-3sqrt(3)ir^2)(2q+3sqrt(3)ir^2)#

Since both cofficients are positive and the one in #q# is of degree #2#, this has no simpler polynomial factors with Real coefficients.

We can do something with Complex coefficients.

I will use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a = 2q# and #b=3sqrt(3)ir^2#

#4q^2+27r^4#

#=(2q)^2+(sqrt(27)r^2)^2#

#=(2q)^2+(3sqrt(3)r^2)^2#

#=(2q)^2-(3sqrt(3)ir^2)^2#

#=(2q-3sqrt(3)ir^2)(2q+3sqrt(3)ir^2)#

This is as far as we can go, even with Complex coefficients, since the degree of the #2q# terms is #1#.